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HW05energy:7-CQ6, 7-CQ10, 7-6, 7-14, 7-15, 7-24, 7-25, 7-33, 7-40, 7-42, 7-56, 7-62. Possible quiz questions:7-CQ6, 7-15

24, 7-25, 7-33, 7-42, 7-56, 7-62.

Chapter 7, Concept Question 6:Figure 7-18 gives the x

component xF of a force that can act on a particle. The

particle begins at rest at x = 0. (a) What is its coordinate

when it has its greatest kinetic energy? (b) What is its

coordinate when it has its greatest speed? (c) What is its

coordinate when it has zero speed? (d) What is the particles

direction of travel after it reaches x = 6 m?

We use the equation,

( ) ;f

i

x

f ix

W dW F x dx K K K (

The particle has an extreme-value of kinetic energy at both x = 3 m and x = 7 m; the x = 3 m has a maximum of kinetic energy, whi

x = 7 m features a minimum in kinetic energy.

Of these, the kinetic energy at x = 3 would be, illustrating the counting up of rectangles and triangles explicitly,

31 1

rectangle triangle 1 2 1 1 1 22 2 2(3 ) (3 ) (2 ) (1 )( ) 2 (1 ) ;K m W m W W m F m F F m F F F m (

1 1triangle rectangle triangle 1 1 2 12 2

3 11 1 1 2 1 2 12 2

(7 ) (7 ) 2 (1 )( ) 2 ( ) ( )(1 ) 2 (3 )

2 ( )

K m W m W W W m F m F F F m K m

F F F F F F m F m

The absolute value of (1.2) is greater than that of (1.3). Since speed is a monotonically-increasing function of kinetic energy,

2 /v K m (positive-root taken for positive-definitespeed), x = 3 m is the position where the speed is greatest.Technically x = 0 m is where the particle has zero speed (in the problem-statement, it is said that the particle is at restat x = 0), but

problem is probably also interested in more than that. By area-counting, you can verify that x = 6 m is also a point where the partic

at rest (notice: 1 12(7 ) ( )K m F m , so it remains only to subtract off the negative triangle to bring the work done on the particle

back up to,

1 1 11 1 12 2 2

(6 ) (7 ) ( ) ( ) ( ) 0K m K m F m F m F m (

The direction of travel is about to reverse at x = 6 m, since it has spent up until this position slowing down the positive velocity it ha

accumulated in going from x = 0 m to x = 3 m; therefore, the direction of travel is negative.

Chapter 7, Concept Question 10:A glob of slime is launched

or dropped from the edge of a cliff. Which of the graphs in

Fig. 7-21 could possibly show how the kinetic energy of the

glob changes during its flight?

The glob necessarily flies through the air freely, under the sole influence of gravity; therefore, kinetic energy should be a

monotonically increasing function of time. We know velocity as a function of time since weve done free -fall problems to death. W

write,

2 2 21 1 10 0 0 0 02 2 2

( ) ( ) ( ) ( ) ;y xv v at v gj t v gj t K K t mv t mv v m v gt v (

Since the particle is either dropped "or" launched, the kinetic energy could be a graph that has a nonzero kinetic energy for all time

The x-component of velocity is constant, and thus contributes to the kinetic energy. The minimum of kinetic energy would thus

correspond to where the kinetic energy is due solely to the x-velocity. Hence, (e) is not permitted because the particle must have an

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velocity, because / 0dK dt at 0t implies the particle was thrown, necessitating a nonzero minimum of kinetic energy.Moreover: ifthe slope /dK dtis positive at 0t , this immediately implies the particle is dropped rather than thrown, implying tthere cantbe a minimum to the kinetic energy. This would permit choices f and g. In choice- h, the dimple in the graph isnt

supposed to be there (according to the class-textbook; the Kof (h) is nonzero at 0t , implying it was thrown; however, if it was

thrown, then it must have / 0dK dt at 0t , which it does not. Therefore, this is the special case of the glob of slime beingthrown horizontally. In short,

0 0 0 0

0 0

( )( )( )( )( );

( ) [slime drops] 0; ( ) [slime thrown] ( , );

( ) [slime thrown horizontally] ( ,0);

x y

x

a b c d e

f v g v v v

h v v

(1

Chapter 7, problem 6:A force aF is applied to a bead as the

bead is moved along a straight wire through displacement

+5.0 cm. The magnitudea

F is set at a certain value, but the

angle between aF and the bead's displacement can be

chosen. The figure gives the work W done by on the bead for a

range of values; W0= 25 J. How much work is done bya

F if

is (a) 64 and (b) 147?

We do not know the magnitude of the force; get it from the part of the plot where 0 . Using 25 JW when 0 and

5.0 cmd . This yields the magnitude ofaF :

225cos 5.0 10cos 0.050 cos0

a a a

W JW F d F d F N

d m

(

(a) We no longer need the plot. For 64 , we have 2cos (5.0 10 N)(0.050 m) cos 64 11 J.aW F d

(b) Repeat what you did in (a). For 147 , we have 2cos (5.0 10 N)(0.050 m)cos147 21 J.aW F d

Chapter 7, Problem 14:The figure here shows an overhead

view of three horizontal forces acting on a cargo canister that

was initially stationary but that now moves across a

frictionless floor. The force magnitudes are F1= 2.9 N, F2=

3.7 N, and F3= 10 N, and the indicated angles are 2= 52

and 3= 31 . What is the net work done on the canister by the

three forces during the first 4.5 m of displacement?

The forces at work need to be written in rectangular form,

1 1 1 1 2 2 2 2 3 3 3 3

1 2 3 1 2 2 3 3

(cos cos ); (cos cos ); (cos cos );

2.9 ; 3.7 ; 10 ; 180 ; 180 90 218 ; 31 ;

F i j F i j F i j

F N F N F N

F F F (

The net force acting upon the cargo-canister is,

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1 2 3

in 1st quadrant 0 90force in 1st quadrant

2 2 1

10cos31 3.7cos218 2.9cos180 10sin31 3.7sin 218 0

2.872 2.756 2.872 2.756 2.872 tan 3.98 46.2 ;

2.756

i j N

x y N N

F F F F

(

This means the displacement is,

4.5 46.2 4.5 (cos 46.2 sin 46.2 ) 4.5 0.692 0.722 ;m m i j i j m d

(1

Take the dot product of this displacement with the force-sum 2.756 2.872x y N F , of (1.9), and you get,

2.756 2.872 4.5 0.692 0.722

4.5 (2.756)(0.692) (2.872)(0.722) 17.91 ;

net net W i j N i j m

N m J

F d F d (1

Chapter 7, problem 15 (|| 13):Figure 7-27 shows three

forces applied to a trunk that moves leftward by 3.00 m over a

frictionless floor. The force magnitudes are F1=5.00N,

F2=9.00N, and F3=3.00N, and 60 . During thedisplacement, (a) what is the net work done on the trunk by the

three forces and (b) does the kinetic energy of the trunk

increase or decrease?

Resolving all of the abovementioned vectors into components,

1 2 3 3.00 ; 5.00 ; 9.00 cos60 sin60 ; 3.00 ;x m x N N x y Ny d F F F (1

Thus, the work done by each force is,

1 1 3

2 2

( 3.00 )( 5.00 ) 15.0 ; ( 3.00 )( 3.00 ) 9 0 0 ;

1 3 ( 3.00 )(9.00 )(cos60 sin60 ) 27.0 ( 1 0) 13.5 ;

2 2

W m N x x J W m N x y J

W m N x x x y J J

F d

F d (1

Net work can be computed as the sum of the contributions of (1.13); which can be shown to be the same as if we computed the vec

resultant (harder) of the sum of the forces1 2 3

F F F F , 1 2 3 1 2 3 1 2 3 15.0 13.5 0 1.5W W W W J J F d F d F d F F F d F d (1

Since 1 2W W , the box is accelerated in the direction of 1F . The box is moving in the left direction, the sam

direction as 1F . Therefore, the kinetic energy increases.

Chapter 7, Problem 24:In the figure, a horizontal force aF of

magnitude 20.0 N is applied to a 3.00 kg book as the book

slides a distance d = 0.500 m up a frictionless ramp at angle

= 30.0. The book begins with zero kinetic energy. What is its

speed at the end of the displacement?

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(a) Find the net work done on the textbook: The net work is the net-force dotted with the displacement. Using incline-plane

coordinates, the component of aF that lies along the inline is cosaF , while the component of gravity lying along the incline is

sinmg (easily recalled from the force-problems we solved). Then,

cos sin

20.0 cos30.0 (9.81)(3.00)sin 30.0 0.500 1.31 ;

x x aW F d F mg d

x N m J

F d (1.

(b) Eq. 7-10 (along with Eq. 7-1) then leads to

212 1

2

cos sin 2(1.31 )cos sin 0.935 ;

3.00

aa

F d mgd J mK W mv F d mgd v

m kg s

(1

Chapter 7, Problem 25:In the figure, a m = 0.250 kg block

of cheese lies on the floor of a M = 900 kg elevator cab that is

being pulled (accelerated!) upward by a cable through

distance d = 2.40 m. If the normal force on the block from the

floor has constant magnitude FN= 3.00 N, how much work is

done on the cab by the force from the cable?

The acceleration of the elevator is unknown, but can be found out from the mass of the cheese, and the normal force the floor of the

elevator is exerting upon it. The net upward force is given by applying Newtons 2ndLaw applied to both the cheese and the elevato

car, in which the acceleration of the cheese equals the acceleration of the elevator (conservation of string again). Let Fbe the fothat accelerates the cheese-elevator system upwards. Then,

( ) ; ; ;m M SYS SYS m M F ma Ma m M a M a M m M a a a (1In (1.17), the force-sum is given by,

The solution manu WRONGal is to put in the force-sum as it does! Considering M+m as the system, has a Newton's 3rd Law pair.

applied gravitational( ) ;

force force

N NF F

F F mg Mg F m M g

(1

We alsoapply Newtons 2ndLaw to the cheese alone: the twoforces acting upon the cheese are NF and mg , which is such that

cheese has a positive acceleration of ma a ,

2solve for a 23.00 N (0.250 kg)(9.80 m/s ) 2.20 m/s ;

0.250 kg

Nm N

F mgF F mg ma a

m

(1

Thus the force from the cable is given by equating (1.17) and (1.18),

solve for( ) ( ) ( )( ) = ;F N NSYS SYS N

F mg F M mm M a F m M g F m M a g M g M F

m m m

The work done by the cable on the cab, subsequently, is,

4 4 1 (1.80 10 N)(2.40 m) 2.59 10 J ;

N N

M m MW Fd j j F d F d

m m

F d (1

N.B.: I just want to point out that the picture of the elevator and the block of cheese is NOT a schematic. Why not? Because no forcare labeled, neither is a coordinate system indicated, or units, etc. Thats what distinguishes a picture from a schematic. T he above

illustration of a yellow block in an elevator is about as much a schematic as various pictures of Grumpy Cat are.

Chapter 7, Problem 33:The block in the figure lies on a

horizontal frictionless surface, and the spring constant is k =

50 N/m. Initially, the spring is at its relaxed length and the

block is stationary at position x = 0. Then an applied force

with a constant magnitude of F = 3.0 N pulls the block in the

positive direction of the x axis, stretching the spring until the

block stops. When that stopping point is reached, what are (a)

the position of the block, (b) the work that has been done o

the block by the applied force, and (c) the work that has be

done on the block by the spring force? During the block's

displacement, what are (d) the block's position when its kin

energy is maximum and (e) the value of that maximum kine

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energy?

The work done upon the spring when the spring is the system, SW , is negative-definite, 0SW , since the springstoresenergy.

Considering the spring as the system, we compute a negative work.

2 212

( ) ( ) ( );f f

i i

x x

S S S f ix x

W dW F x dx kx dx k x x (1

The work done upon the system when the agent of the force is the system is the negativeof this work, A SW W ,

;AW Fd F d (1

Putting (1.22) and (1.23), we compute a distance of,

solve for2 2 2 2 21 1 12 2 2

2 2(3.0 )( ) ( 0 ) 0.1250

fx d

A S f i Nm

F NW W Fd k x x k d Fd kd d mk

The applied work, 0AW , and the spring-work, S AW W , each are,

(3.0 )(0.12 ) 0.36 ;A SW Fd N m J W (1

Kinetic energy is a function of position, and is zero at 0x and at 2 /x d F k , as,

*

2 * *12

3.0( ) ( ) ( ) ( ) 0 0.060 ;

50A S

x x

dK FK K x W x W x Fx kx F kx x m m

dx k (1

This value of maximum kinetic energy is,

2 2 2

* * *2 21 1max 2 2 1 1 (3.0)( ) 0.090 ;

2 2 2(50)F F FK K x Fx kx F k F J Jk k k k k

(1

chapter 7, problem 40 (|| 7.42):A can of sardines is made to move along an x-axis from x = 0.25 m to x = 1.25 m by a force wit

magnitude given by2

0( ) xF F x F e

, with x in meters, 0 1F N , and 214

m . (Here exp is the exponential function.) Ho

much work is done on the can by the force?Solution:Using the numerical integral22.5

0.50.425

ue du and integration by

substitution (this is a very good exercise) with the change of variables u x ,

212 2 2 2

2

1.25 4 1.25 2.5(4 )

( ) 0 00 0 0

0.25 0.5

4 0.25

2

2

( )

10.210. 0.21 0.21 ;

4425

f f

m

i i

x xmx

x x u u

m x x

F Fd xW F e dx F e dx F e e du e du

NN m N m J

m

(1

Chapter 7, Problem 40 (modified!):Acan of sardines is made to move along an x axis from x = 0.23 m to x = 1.11 m by a force

with a magnitude given by F = exp(7x), with x in meters, -7 carrying units of m -1, and F in newtons. (Here exp is the exponential

function.) How much work is done on the can by the force? First, to get the units right, rewrite this as 0kxF F e , where 0 1F N

and17k m

(inverse-meters, often found in the argument of an exponential function). Then,

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1 1

1.111.11 1.11

00

0.23 0.230.23

7 1.11 7 0.23 2 2

1 1

( ) ( )

12.85 10 2.85 10

7

x mkx

m mkx

x x m mx m

m m m m

F eW dW d F x x F x dx F e dx

k

N Ne e J

m m

(1

In the 2ndstep, although the Product Rule tells us ( ) ( ) ( )d F x x dF x x F x dx , we regard x to notbe a function of force

(that would be sillya coordinate-dependent force), and so ( ) 0x dF x , which would, on definiteintegration, disappear, and

indefiniteintegration become an unphysical constant (constant shifts to energy have absolutely no physical interpretation).

Chapter 7, Problem 42:The figure shows a cord attached to

a cart that can slide along a frictionless horizontal rail

aligned along an x axis. The left end of the cord is pulled over

a pulley, of negligible mass and friction and at cord height h

= 1.20 m, so the cart slides from x1 = 3.00 m to x2 = 1.00 m.

During the move, the tension in the cord is a constant 25.0 N.

What is the change in the kinetic energy of the cart during

move?

The work-energy theorem is used, but one must use the identityadjacent

hypotenusecos upon cosW F d F d (in which I

illustrated the angle in the schematic), and eliminate the trig-functions in favour of the dimensions illustrated above. Then, one u

the antiderivative2 1

2

0

1xx

x xd x

, where x is the constant of integration, and write,

2 2 21 1 1

Solution manual is again: the antiderivative is only fortuitously/incidentally equal to force times displacement.

The force

2 2 2 2

2 12 2

wrong

( ) ( )cos ( )x x x

x x x

xW dW F x dx T x dx T dx T x h x h

x h

2 2 2 22 1varies, and blindly mutliplying T by is technically incorrect.

2 2 2 2(25.0 ) (1.00 ) (1.20 ) (3.00 ) (1.20 ) 41.7 41.7 ;

d x h x h

N m m m m J J

(1

Chapter 7, Problem 56 SN:An initially stationary object of mass m accelerates horizontally and uniformly to a speed vfduring a

time interval of length t.this part of the problem statement says,

0( ) ( ) 0 0 ( / ) 0 ( ) 0 ( ) ;fvv

ft tv v t v at v t at F m t t t v (1

(a) In this interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that

force (b) at the end of the interval and (c) at the end of the first half of the interval?

2 2 2 2 2 2 21 1 1 1 1

02 2 2 2 2

f i f i f i f fW K K K mv mv m v v m v mv (1

Instantaneous power due to this force is force times velocity,

2 2

12( ) ( ) ( ) ( ) ( ) ; ( ) ( )( ) ;

2 2f f f f ff f f

v mv v v mvvP P t Fv t P t Fv m v m v P t m

t t t t t

(1

Chapter 7, Problem 62 SN:A block of mass m is dropped onto a relaxed vertical spring with spring constant k (see the figure). Th

block becomes attached to the spring and compresses the spring a distance d before momentarily stopping. While the spring is bein

compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of th

block just before it hits the spring? (Assume that friction is negligible.) State your answers in terms of the given variables and g.

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There are two stages of motion: (1) mass m falls and hits the spring, under the influence of gravity, and (2) the mass m compresses

spring to a distance 0 d d , under the influence of bothgravity and spring-force. Treating the mass-m as the system, we have

2 212

212 2 2 2 21 1

2 2

( )( ) [independent of up/down being / ];

[force is opposite (negative) direction of gravity, while d 0, identically];

2( ) 2( )( ) (0 )

g

s

g s

f i i g s i

W m g d mgd

W kd

W W kd mgd K m v v m v W W v

m

;

m

(1